Solve for the real scalar value of $k$, if $z =\frac{2}{1+ki} -\frac{i}{k-i}$ and $z$ lies on the line, $y= 2x$

Question

The complex number $z$ is given by $\frac{2}{1+ki} - \frac{i}{k-i}$. If it is given that $z$ lies on the line, $y = 2x$, find the value of the real scalar $k$.

So far this is what I have come up with : (See attached photo)

I am stuck at this point. I think that, because the question asks for the value of the real scalar $k$, I should move forward with only the "$x$" part of the coordinate system, but I do not know what to do with it.

Also, because it asks for the scalar value, I am wondering if I need to find the modulus of anything.

Any help would be appreciated.

Answer 1

The equations derived in the picture for $(x,y)$ are wrong from line 4 onward. The corrected ones are

$$z=\frac{3k-3i}{2k-i(1-k^2)}\times\frac{2k+i(1-k^2)}{2k+i(1-k^2)}=\frac{3k^2+3+i(-3k^3-3k)}{4k^2+(1-k^2)^2}$$

We have $y=2x$ and $x = \frac{3k^2+3}{4k^2+(1-k^2)^2}$ and $y = \frac{-3k^3-3k}{4k^2+(1-k^2)^2}$ therefore

$$\frac{-3k^3-3k}{4k^2+(1-k^2)^2}=2\times \frac{3k^2+3}{4k^2+(1-k^2)^2} \Rightarrow -3k^3-3k=6k^2+6 \Rightarrow k^3+2k^2+k+2=0\\ \Rightarrow k^2(k+2)+(k+2)=0 \Rightarrow(k+2)(k^2+1)=0 \Rightarrow k=-2,i,-i $$

That's all the possible answers, of course only $k=-2$ is real and obviously $k=i,-i$ are not in the domain of $z=\frac{1}{1+ki}-\frac{i}{k-i}$ the original problem.

Answer 2

$$z =\frac{2}{1+ki}\cdot \frac{1-ki}{1-ki} -\frac{i}{k-i}\cdot\frac{k+i}{k+i},$$

$$z=\frac{3}{1+k^2} - i \frac{3k}{1+k^2}.$$

So, setting the imaginary part equal to twice the real part $$-3k=6$$ and thus $$k=-2.$$